Question

Please help with graphing a quadratic function. y=(x-1)^2 Got stuck at y=(x-1)(x-1) and p=1, q=1 Which doesn't work, because you can't fix a vertex with one point on a graph.

Answer 1

What is Alberbra?

I don't know what your p and q are upposed to be.

Define a new variable x' = x-1. Your equation becomes
y = x'^2

The vertex of your parabola is at x=1, y=0.

That is one point of the graph. Other points are easily plotted.

The parameter p is often used to designate the distance from the vertex to the focus. In tis case, p = 1/4, since the standard parabola formula is y = 4px^2 , with p being the focal distance.

Answer 2

The standard quadratic is
y=a(x-h)²+k

where (h,k) is the vertex, and x=h is the axis of symmetry.
The zeroes (p,q) are at -h±√(k/a).
If a>0, the quadratic is concave upwards, i.e. the vertex is at the minimum.
If a<0, the graph is concave downwards, or the vertex is at the maximum.


In the case of y=(x-1)²,
a=1, h=1, k=0
The zeroes are coincident because k=0, and coincide with the vertex. This also means that the graph is tangent to the x-axis.
I will leave it to you to figure out the concavity and the maximum/minimum.
Post if you need more information.